1. ## Mean Value Theorem

Mean value theorem states that for continuous function f on [a, b] and differentiable on (a, b), there is at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}.$

Consider the function $f(x) = x^3.$
f'(x) = 0 at x = 0. However, there is no interval [a , b] that satisfies mean value theorem (since f(b) > f(a) for all b > a).

Why does the Mean value theorem fail here?

2. Originally Posted by Rozaline
Mean value theorem states that for continuous function f on [a, b] and differentiable on (a, b), there is at least one point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b-a}.$

Consider the function $f(x) = x^3.$
f'(x) = 0 at x = 0. However, there is no interval [a , b] that satisfies mean value theorem (since f(b) > f(a) for all b > a).

Why does the Mean value theorem fail here?

It doesn't: the logical direction is ==> , not <== : for any a < b in the interval we have [f(b) - f(a)]/[b-a] = f'(c), for some c in (a,b), but this doesn't meant that ANY value f'(c) of the derivative has to be obtained this way.

Tonio